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Unformatted text preview: 1 7. Dynamical System 1order Firstorder system: The differential equation is Solution: Take Laplace Transform: If u(t) is a unit step, ) t ( u ) t ( y dt ) t ( dy a = + ∴ 1 as ) s ( U ay ) s ( Y ) s ( U ) s ( Y ay ) s ( asY o o + + = ∴ = + 1 as ay ) s ( Y s 1 o + + = ∴ 2 1order systems (cont.) Let y o = 0, i.e. no initial condition then {} a t a 1 e 1 s 1 L L ) t ( y s 1 s 1 Y(s) a s 1 lim B , 1 1 as 1 lim A 1 as B s A ) 1 as ( s 1 ) s ( Y a 1 1 s 1 1 a 1 s s → → = + = + = ∴ = = = + = + + = + = τ 62.3% 1 y(t) t 3 1order systems – time constant The timeconstant is defined as the time taken to reach the full value if the initial rate is maintained. Initial rate is Therefore the time to reach 1 takes a seconds; and at t = a And the time constant is a seconds. t dt dy = 63.2% i.e. , 632 . e 1 ) t ( y 1 = = a 1e1 1 y(t) t ( 29 a 1 t e e 1 dt d dt dy a t a t a 1 = = = = 4 Dynamical Systems Behavior (2order) We look at a singledegreeof freedom (sdof) system first. Usually we represent this system by a springmass damper system as shown. m c k x(t) f(t) • The reason we call it sdof is: only one variable, i.e....
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 Spring '08
 LUI
 Laplace, Trigraph, damped natural frequency, unit step

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